Buffon's needle with different distances I've solved problem with Buffon's needle with : l - length, d - distance. It gave me $\mathbb{P}(A) = \frac{l_2(A)}{l_2(\Omega)} = \frac{2l}{\pi d}$. So if I drop 2 cm long needle on infinite, parallel lines 4 cm distant, probability is equal to $\frac{1}{\pi}$. There is my question. What if distances between every two lines aren't equal? What if distances are alternately 3 cm and 5 cm? Would it be two cases with probability of 1st = $\frac{4}{3 \pi}$ and 2nd = $ \frac{4}{5\pi}$? How would final probability look like?

Define events:

\begin{eqnarray*} A &=& \text{The needle crosses a line} \\\ B &=& \text{The needle centre is in a 5cm strip.} \\\ \end{eqnarray*}

Clearly, $P(B) = \frac{5}{3+5}$. Also, once $B$ occurs, the probability of $A$ occurring is the same as if _all_ lines were 5cm apart. So using conditional probability,

\begin{eqnarray*} P(A) &=& P(A\mid B)P(B) + P(A\mid B^C)P(B^C) \\\ &=& \dfrac{4}{5\pi}\cdot \dfrac{5}{8} + \dfrac{4}{3\pi}\cdot \dfrac{3}{8} \\\ &=& \dfrac{1}{\pi}. \end{eqnarray*}

This is the same result as when all lines are 4cm apart. This makes sense because the fact that every second line is moved doesn't affect the overall probability - as long as no two lines get closer to each other than the length of the needle - because the needle is equally likely to land anywhere.